Chemistry · Structure Of Atom · NEET
An s orbital is spherical (a round ball around the nucleus). A p orbital is dumbbell-shaped: two lobes on opposite sides of the nucleus, with a gap in the middle where the electron is never found. A d orbital normally has four lobes (a double-dumbbell or clover shape). The shape is decided by the azimuthal quantum number l: l=0 gives s (sphere), l=1 gives p (dumbbell), l=2 gives d (four lobes). For NEET, remember l decides the SHAPE.
An s orbital has l=0, so it has no angular node (no flat plane where the electron is missing). This makes the chance of finding the electron the same in every direction, which gives a sphere. A p orbital has l=1, so it has one angular node (a flat plane through the nucleus). On that plane the chance of finding the electron is zero, so the shape breaks into two lobes on either side, giving a dumbbell. More angular nodes means a shape with more lobes.
There are five d orbitals. Two of them, d(z²) and d(x²-y²), have their lobes pointing ALONG the x, y, z axes (this pair is called the e_g set). The other three, d(xy), d(yz) and d(xz), have their lobes lying BETWEEN the axes (this trio is called the t_2g set). NEET 2016 asked exactly this: the pair with electron density along the axes is d(z²) and d(x²-y²).
Four of the five d orbitals have the same clover shape (four lobes), just pointing in different directions. The d(z²) orbital is the odd one out: it has two big lobes along the z-axis plus a small doughnut-shaped ring (a torus) around the middle in the xy-plane. Its shape looks different but it still belongs to the same set of five d orbitals with equal energy. Its magnetic quantum number m is 0.
No. The shape is only a boundary surface: the region where there is about a 90 percent chance of finding the electron. An orbital is NOT a fixed path like a Bohr orbit. Because of Heisenberg's uncertainty principle, we cannot know the exact path. The shape just shows where the electron is most likely to be, based on ψ² (probability density).
An s orbital has 0 angular nodes and no lobes (it is one sphere). A p orbital has 1 angular node and 2 lobes (dumbbell). A d orbital has 2 angular nodes and usually 4 lobes. The number of angular nodes equals l. For NEET, angular nodes = l, so s=0, p=1, d=2, f=3.
Which of the following pairs of d-orbitals will have electron density along the axes?
Match List-I (Quantum Number) with List-II (Information provided). (A) m_l (B) m_s (C) l (D) n. List-II: (I) shape of orbital, (II) size of orbital, (III) orientation of orbital, (IV) orientation of spin of electrons.
Which one is a wrong statement?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The azimuthal quantum number l decides the shape. l=0 gives a spherical s orbital, l=1 gives a dumbbell p orbital, and l=2 gives a four-lobed d orbital. The principal quantum number n decides size, and m_l decides orientation.
There are three p orbitals (p_x, p_y, p_z) and five d orbitals (d(xy), d(yz), d(xz), d(x²-y²), d(z²)) in each subshell. The number of orbitals equals 2l+1.
Yes, all s orbitals are spherical. But the size grows with n, so 4s > 3s > 2s > 1s. Higher s orbitals also have more radial nodes inside, but the outer boundary shape stays a sphere.
Four d orbitals have the same four-lobe clover shape. The d(z²) orbital is special: it has two lobes along the z-axis plus a doughnut-shaped ring around the centre. All five still have equal energy in a free atom.
No. A shape is just the region where the electron is very likely to be (about 90 percent chance). It is not a fixed path. By Heisenberg's uncertainty principle, the exact path of an electron cannot be known.
